Introductory Analysis 2 Spring 2010 Exam 1 February 11, 2015

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1 Introductory Analysis 2 Spring 21 Exam 1 February 11, 215 Instructions: You may use any result from Chapter 2 of Royden s textbook, or from the first four chapters of Pugh s textbook, or anything seen in class. Anything more advanced needs to be justified (and probably should not be used). You may also use results from previous homework. 1. Let g : [, 1] R be continuous. Show there exists a unique continuous function f : [, 1] R such that f(x) = g(x) + e (x t)2 f(t) dt. Hint: All you need to do here are some simple computations, and quoting past results correctly. Make sure you quote them correctly. Something that may or may not be useful (depending on how you decide to solve the problem) is: If φ : [a, b] R is continuous, φ(x) for all x [a, b] and φ is NOT identically zero, then b φ(t) dt >. a Solution. The idea in solving this exercise is essentially the same as in Picard s Theorem. The space M = C([, 1]) = {f : [, 1] R} is (as seen in part 1) a complete metric space with the metric defined by d(f, g) = f g, where Given g M, Define L : M M by L(f)(x) = g(x) + f = max x1 f(x). e (x t)2 f(t) dt. It is clear that Lf : [, 1] [, 1] is continuous; that is Lf M if f M so that we do indeed have L mapping from M into M. If f, h M, then for x [, 1], L(f)(x) L(h)(x) = e (x t)2 f(t) dt e (x t)2 h(t) dt e (x t)2 f(t) h(t) dt ( ) ( ) e (x t)2 dt = e t2 dt f g ( 1 ) e t2 dt f g. This being true for all x [, 1], we proved L(f) L(h) α f h ; i.e., d(l(f), L(h)) αd(f, h) for all f, h M, where α = 1 dt. We notice that < α < 1. In fact, e t2 e t2 < 1 for all t (, 1] so that! e t2 > for all t (, 1], hence 1 α = 1 (1 e t2 ) dt >. It follows that L is a contraction from M to M. since (as mentioned) M is complete, L has a fixed point; there is f M such that f = Lf.

2 2. If I = (a, b) is an open, bounded, interval in R, define λ(i) = e b e a. If A R define µ (A) = inf{ λ(i n ) : {I n } n N ranges over all families of bounded open intervals s.t. A I n }. Prove that µ is an outer measure in R; that is, prove: (a) µ ( ) =. (b) A B implies µ (A) µ (B). (c) µ ( A n) µ (A n ) for all countable families A 1, A 2,... of subsets of R. Solution. The proof of all these properties is identical to the proof that the outer measure m defined in the textbook is indeed an outer measure. I allowed the covering intervals to be empty, which simplifies the proof. However, that is not necessary. Proof that µ ( ) =. Since x e x is continuous, given ϵ >, n N we can find a n, b n R, a n < b n such that e bn e an < ϵ/2 n+1. Setting I n = (a n, b n ) we have λ(i n ) < ϵ/2 n+1. Now I n, hence µ ϵ ( ) λ(i n ) < = ϵ. 2n+1 Since ϵ > was arbitrary, we are done. Proof that µ is monotone. Assume A B R. If {I n } is any family of open intervals covering B, it also covers A hence µ (A) λ(i n ). This implies that µ (A) is all numbers in the set of which µ (B) is the greatest lower bound. Thus µ (A) µ (B). Proof of σ-subadditivity. Assume A, A 1, A 2,... are subsets of R and A A n. Let ϵ > be given. For each n N there is a family {I nk } k N of open and bounded intervals such that A n I nk and λ(i nk) < µ (A n ) + ϵ 2. Then {I n+1 nk } n,k N is a covering of A by a countable family of open intervals,hence µ (A) k,n N λ(i nk ) = λ(i nk ) < Since ϵ > is arbitrary, we are done. ( µ (A n ) + ϵ 2 n+1 ) = Bonus: Is µ (I) = λ (I) if I is a bounded open interval? µ (A n )+ϵ. The answer is yes, the proof is similar (slightly easier because you have to deal only with open, bounded intervals) than the proof that the outer measure of an interval is its length. The fact that µ (I) λ(i) is trivial by the definition. For the converse inequality, assume I = (a, b) I n, where each I n is an open bounded interval. One needs to prove that λ(i ) λ(i n). One proceeds as for Lebesgue outer measure but 2

3 with a bit of care. The idea is identical, but shrinking an interval by ϵ > does not decrease its λ value by ϵ. We need an ϵ and a δ. We can argue as follows. Because I is open we can assume I n I for all n. In fact, otherwise replace I n by I n I; I n I is againa an open interval and because x e x is strictly increasing, λ(i n I) λ(i n ), so if we prove λ(i ) λ(i n I), then λ(i ) λ(i n) follows. It may be convenient to define f : R R by f(x) = e x. Because it is continuous, it is uniformly continuous on bounded sets; given ϵ > there is δ > such that x y δ; x, y I implies f(x) f(y) < ϵ/2. Let J = [a+δ, b δ] so that J is a closed and bounded interval, J I. By compactness, there is N N such that I 1,..., I N cover J. We relabel these intervals as follows as in Royden. There is one of these intervals containing a + δ. Relabel it as I 1 = (a 1, b 1 ). If b 1 < b δ we are done. If not, there is another one of this finite set of intervals containing b 2. Relabel it as I 2 = (a 2, b 2 ). And so forth. The process has to end, eventually one of the intervals contains b δ, since we only have a finite number of intervals. So we have for some m N a 1 < a + δ < a 2 < b 1 < b 2 < a 3 < < a m < b δ < b m. Now, as in Royden (p. 32), but adding the function f. Since f is increasing, it preserves order relations. Notice that (f(a m ) f(b m 1 )) + (f(a m 1 ) f(b m 2 )) + (f(a 2 ) f(b 1 )) < because b k (a k+1, b k+1 ); subtracting this expressions increases quantities. So, as in Royden, but with an f around points, f (b δ) f (a + δ) < f(b m ) f(a m ) < f(b m ) (f(a m ) f(b m 1 )) (f(a 2 ) f(b 1 )) f(a 1 ) = (f(b m ) f(a m )) + (f(b m 1 ) f(a m 1 ) + + (f(b 1 ) f(a 1 )) m = λ(i k ) This proves that f(b δ) f(a + δ) < λ(i n ), where now the I n s are again the original I n s. By choice of δ, f(b δ) > f(b) ϵ/2, f(a + δ) < f(a) + ϵ/2 so that λ(i) = f(b) f(a) < f(b δ) f(a + δ) + ϵ < m λ(i k ) + ϵ. Since ϵ > is arbitrary, this proves λ(i) m λ(i k); since {I n } was an arbitrary covering of I by open, bounded intervals, this proves λ(i) µ (I). 3. Let E be a measurable subset of R. Prove: m(e) = sup{m(c) : C is a compact subset of E}. Solution. Maybe I should have made this easier by adding the assumption that E is bounded. Then it is an immediate consequence of Theorem 11 in Royden, since closed bounded sets are compact. For an unbounded E there are several ways to proceed. one way is to prove it first for a 3

4 bounded set. So let s do this, assume first E is bounded. Let ϵ >. By Theorem 11 (p.4) in Royden, there is a closed sibset F of E such that m(e\f ) < ϵ. Since bounded sets have finite measure, we see that m(e) m(f ) < ϵ, hence m(e) < m(f ) + ϵ sup{m(c) : C closed,c E} + ϵ. Since ϵ > was arbitrary we see that m(e) is the sup of the measure of all included closed sets; since the opposite inequality is obvious we have that m(e) = sup{m(c) : C is a closed subset of E} = sup{m(c) : C is a compact subset of E} since compact and closed are the same thing for bounded subsets of R. For the general case, let E n = E [ n, n]. There is then a compact set C n E n such that m(e n ) < m(c n )+1/n. Here is the place where we have to be a little bit careful. So far, to say something like m(a) < m(b) + ϵ for measurable sets A, B, B A, as long as A has finite measure, is equivalent to m(a\b) < ϵ. But if the measures get infinite, things change. Back to E n, C n. We have E 1 E 2 E 3 and E = E n, thus m(e) = lim n m(e n ). We now have unfortunately to consider two cases. Case 1: m(e) <. Given ϵ >, there is then n such that m(e n ) + ϵ/2 > m(e) and such that 1/n < ϵ/2. Then m(e) < m(e n ) + ϵ 2 < m(c n) + 1 n + ϵ 2 < m(c n) + ϵ. We proved that for every ϵ >, E contains a compact subset with measure > m(e) ϵ; the conclusion follows. Case 2. m(e) =. In this case lim n m(e n ) = thus also lim n m(c n ) =. Thus for every R > there is n such that m(c n ) > R; in other words, for every R >, E contains a compact subset of measure > R. It follows that sup{m(c) : C is a compact subset of E} = = m(e). 4. Prove or disprove: There exists a closed subset of the interval [, 1], consisting exclusively of irrational numbers, having positive measure. That is, can there exist F closed, F [, 1]\Q, m(f ) >? Solution. Since the set of irrational numbers is a measurable set of infinite measure, by the previous exercise it contains compact, hence closed, subsets of arbitrarily large measure. So there certainly exists such a set F. Many such sets, even compact such sets. 5. Let E R, assume m (E) <. (a) Show there exists an F σ set A and a G δ set U such that A E U and m(a) = m (E) = m(u). Solution. This is exercise 18 (p. 43) in Royden. Half of it is false. But first we first show the existence of the G δ set. One doesn t really need m (E) < for this part; if m (E) = we can take U = R. But the proof for the case m (E) < is a bit more involved. So assume m (E) <. Let ϵ > be given. There exists then a family {I n } of bounded open intervals such that E I n and 4

5 m (E) l(i n) < m (E) + ϵ. Setting U ϵ = I n, U ϵ is open, E U ϵ and The set U = m (E) m(u ϵ ) m(i n ) = l(i n ) < m (E) + ϵ. U 1/k is a G δ set, E U and since m(u) m(u k ) < m (E) + 1 k for all k, we see that m(u) m (E); since E U we also have m (E) m (U) = m(u). This takes care of the existence of the G δ set. But the F σ set might not exist. I will refer to my answers to Homework 3, Exercise 1. The same argument I use there proves that if E is a measurable set of positive finite measure, if A = C E (defined as in Royden), then m (A) > (has to be since A is not measurable, m (A) < (because E A and m(a) < ), and every measurable subset (in particular every F σ subset) of A has measure. In other words, there is NO F σ subset of A with m(f σ ) = m (A). 6. Show there is a continuous, strictly increasing, function on the interval [, 1] that maps a set of positive measure onto a set of measure. Hint: What goes up must come down. problem seems hard, invert it. Or as Abel used to say, if a Solution. If the problem had stated: Show there is a continuous, strictly increasing, function on the interval [, 2] (instead of [, 1]) that maps a set of positive measure onto a set of measure, it would have been a trivial consequence of Proposition 21 (p.52) of Royden. In fact, by Proposition 21, the function ψ : [, 1] [, 2] defined there is onto, strictly increasing, continuous and maps C onto a set of positive measure. Since ψ is strictly increasing and onto [, 2], it is invertible and ψ 1 : [, 2] [, 1] is strictly increasing and onto. Moreover, ψ 1 is continuous, since ψ is continuous and [, 1] is compact. Almost finally, if we let E = ψ(c), then by Proposition 11, E is a measurable set of positive measure (in fact, of measure 1). Thus ψ 1 maps a measurable set of positive measure, namely E, onto a set of measure, namely C. The transition to the interval [, 1] is quite easy. Let E be as above. The easiest way to proceed is to say that either E 1 = E [, 1] or E 2 = E [, 2] has positive measure. If m(e 1 ) >, then let f = ψ 1 [,1] ; f is clearly strictly increasing, continuous, and maps the non-null measurable set E 1 onto a subset of the Cantor set C, thus onto a null set. If m(e 1 ) =, so m(e 2 ) >, define g : [, 1] [, 1] by g(x) = ψ 1 (x + 1). Once again it is immediate that g is strictly increasing, continuous. Letting E 3 = E 2 1 (E 2 translate by one to the left), we see that E 3 is a measurable subset f [, 1] of positive measure and g(e 3 ) = ψ 1 (E 2 ) C is a null set. Done. There are other ways one can apply Royden s Proposition. For example one can define the sought for function by x ψ 1 (2x); but then one needs to prove that if E is a subset of [, 2] of positive measure, then {x [, 1] : 2x E} also has positive measure. This is fairly easy, but a bit more time consuming than what I did. 5